User:CNSW

We know for all right triangles that:

$\sin x = \left ( \frac{opp}{hyp} \right )$

$\tan x = \left ( \frac{opp}{adj} \right )$

From the latter we can deduce

$\tan x \times adj = opp$

Through substitution into our first equation, it follows that

$\sin x = \left ( \frac{\tan x \times adj}{hyp} \right )$

And through standard algebra:

$\sin x \times hyp = \tan x \times adj$

$\tan x = \left ( \frac{\sin x \times hyp}{adj} \right )$

Substitute that back into $\sin x = \left ( \frac{\tan x \times adj}{hyp} \right )$